Consider the function $x^{2}$. Its Fréchet derivative should be $2x$ which is confirmed by evaluating the definition of Fréchet derivative $$ \lim_{|h|\rightarrow0}{\frac{(x + h)^2 - x^{2}- 2xh}{|h|}=\lim_{|h|\rightarrow 0}\frac{x^{2} + 2xh +h^{2} - x^{2} - 2xh}{|h|}} = \lim_{|h|\rightarrow 0}\frac{h^{2}}{|h|} = \lim_{|h|\rightarrow 0}\frac{|h|^{2}}{|h|} = \lim_{|h|\rightarrow 0}|h|=0 $$ From the other hand, first order expansion ($f(x + h) = f(x) + f'(x)h + o(|h|)$) must also be true. And for the function in question it is $$ (x + h)^{2} = x^{2} + 2xh + h^{2}. $$ And I have doubts about $h^{2} \in o(|h|)$ because $f(x) \in o(g(x))$ is defined as $\lim_{x \rightarrow+\infty}\frac{f(x)}{g(x)} = 0$. Applying this definition to $h^{2}$ gives $$ \lim_{|h|\rightarrow+\infty}\frac{h^{2}}{|h|} = \lim_{|h|\rightarrow+\infty}\frac{|h|^{2}}{|h|} = \lim_{|h|\rightarrow+\infty} |h| \neq 0. $$ This should mean that $h^{2} \not\in o(|h|)$ but because this function is Fréchet differentiable first order expansion must hold. What am I missing?
My explanation is that in the context of Fréchet derivatives $f(x) \in o(g(x))$ means $\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}=0$ but this is not the usual definition of 'little o' notation, so I am not sure if this explanation is reasonable.